12AX7 Concertina Bias Q's

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cayocosta

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Joined
Jun 3, 2004
Messages
262
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Biasing a 12AX7 concertina phase splitter using a voltage divider from 243V B+. The plate and cathode resistors are both 56k.

Looking for -1.5V grid bias.

Here's what I tried:

2.2M/1M Divider:
Plate:169.9
Grid:71.0
Cathode:73.6
Bias:-2.6V

2M/1M Divider:
Plate: 168.7
Grid:74.6
Cathode: 76.5
Bias:-1.9V

1.8M/1M Divider:
Plate:167.0
Grid:76.7
Cathode:78.3
Bias:-1.6V

The 1.8M/1M divider did the trick - but none of the dividers gave me the voltage I expected to see:

1M / 2.2M+1M = .31 * 243V = 76V
1M / 2.0M+1M = .33 * 243V = 81V
1M / 1.8M+1M = .357 * 243V = 87V

Can someone help me understand what's going on?
 
I am seriously confused. My understanding of a concertina phase splitter is what is sometimes called a split phase inverter. One half of a dual triode with generally equal plate and cathode loads. How is this thing hooked up? Grid resistor seems way small. Do you have a schematic? Sorry to be a pain here.
 
Thomas, the plate and cathode resistors are each 56k. There is no grid resistor. Grid bias is tapped off the B+ with a potential divider as explained above. Works fine - I just don't understand why the voltage of the divider - in the circuit - isn't what it's calculated to be on paper. Obviously I'm missing something.

B+ 243V
------
r1 | | 56k
---+o 12AX7 (one triode)
r2 | | 56k
------
Ground

r1 and r2 are the divider resistors.
 
> none of the dividers gave me the voltage I expected to see:
1M / 2.2M+1M = .31 * 243V = 76V


What is the input resistance of your voltmeter?

10Meg? That's in parallel with your 1Meg resistor, 10Meg||1Meg= 909.09K, so you really have 909.09K in the bottom.

909.09K / (2200K+909.09K) = 0.2924 * 243 = 71.05V
(measured 2.2M/1M Divider: Grid: 71.0)

Looks right to me!

> Looking for -1.5V grid bias.

WHY? That's quite arbitrary.

Design of a split-load starts with specification of the load impedance and drive voltage. And it is really best to convert it to the equivalent single-ouput design.

If you need two grids swing 20V peak with 100K grid resistors, this is similar to a single load of 200K at 40V peak. If you use a 200K plate resistor, you will need at least 80V of DC drop across it to get a 40V peak swing into 200K when the tube cuts off. You then turn the tube to zero-bias and be sure you can get the same swing when the tube saturates. This is often where you discover that a 12AX7 is a very lame driver compared to 12AU7 and kin. You should really start with the single-load equalent and look it up in the RC Amplifier chart: if you can't meet specs, try another tube.

Aside from barely meeting a given spec, you often want to over-design. Actually in most applications of a spliit-load you don't want a LOT of output voltage headroom: you just get into overload troubles in the following stage, and a split-load is very low THD right up to saturation/cutoff. But you may want high current to swamp the effects of capacitance.
 
Thanks PRR.

Yes, my DMM has an input resistance of 10M.

However, factoring in the 10M meter resistance, the remaining 2.0M and 1.8M calculations (above) still don't jibe with actual measurements:

.909M / 2.0M+.909M = .312 * 243V = 76.0V actual 74.6V
.909M / 1.8M+.909M = .335 * 243V = 81.5V actual 76.7V
 
A self-biased split-load PI works as well as the fixed bias version, but has a much higher input impedance.

concertina.gif


In this version, you would infer the grid voltage by reading the voltage at the junction of R1 and R2. Any voltmeter (even with a 10M input impedance) would act as a shunt and give a false reading if applied directly to the grid.

I agree with PRR that the 12AX7 is not the tube to use if you're driving any kind of demanding load.
 
Thank you for clearing up the configuration for me. I am quite slow on the uptake at times.

I have not used anything other than the self biassed configuration posted by NY Dave above. I must also agree that 12AX7 in this application tends to do a much poorer job than it's lower mu, higher current cousins like 12AT7 or 12AU7. I think you will find a 12AU7 in particular a much better driver in nearly every application of a split load inverter.

As far as your voltage measurements on your circuit, I can't see that much error given what you are using to measure with and the circuit you have in front of you. The tollerences of circuit components and test equipment can give quite a bit of variability here. You sure seem in the ball park with your measured results.
 
If you decide to try a 12AU7 self-biased with a 240-250V B+, try 10K for the plate and cathode load resistors, and 1K for the cathode bias resistor. The plate current will be fairly high and it should have plenty of "oomph" to drive typical push-pull output stages.
 
Hey Dave,

To balance the loads equally, would it make sense to up the plate resistor to 57.5K? (Matching the 56k+1.5k on the cathode side.)

Thanks,

Ron
 
Not really. Remember, the source impedance at the cathode is much lower than that at the plate. The relatively miniscule added series resistance at the cathode makes little difference as regards balance.
 
> Just found this:

The cathodyne has always been misunderstood, and we have forgotten most of what they used to know.

That article explains the frequency response issues precisely. But it has a misleading remark at the end. Yes, a cathodyne gives half the swing per output as a single-output stage. BUT if you limit distortion to 5% or so, it has about the same max swing on each output as a one-output stage does!

Consider. A plain grounded-cathode plate-loaded RC amp can give a peak output at 5% distortion of about 20% of supply voltage V+. For 243V B+ you can expect 243V*0.2= 48.6V from a voltage-amp at 5% THD. You can swing further but the wave gets very bent.

The vast negative feedback in the Cathodyne reduces THD by about Mu/2. Taking Mu=20 (12AU7) then at 20% swing (10% per output) the THD is more like 0.5%. If you use a tube with saturated plate resistance much lower than DC load resistors, and keep the AC load high, you can run up around 40% of V+ (20% per output) before THD rises above a few percent.

In practice the Cathodyne is normally fed from a volt-amp eating the same B+ as the Cathodyne. So with 243V B+, the volt-amp will hit 5%THD at about 48V peak, the Cathodyne will convert this to two 48V outputs with only about 1% added THD. Your volt-amp is almost always the limit when looking for large swing with low (under 10%) THD.
 
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